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Let $f$ be a continuous function from $\mathbb{R}$ to $\mathbb{R}$. Does this mean that at each number $a$ there exists an $M$, $N$ $\in \mathbb{N}$ such that $\left|f(a+h) -f(a)\right| < M |h|$ for all $|h|< \frac{1}{N}$?

At first I thought this must be true but now I’m doubting this. I don’t believe this condition implies that $f$ is differentiable since absolute value function satisfies this condition at $x=0$. I think I can prove the condition holds for differentiable functions but am at a loss when assuming just continuity.

skh
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    That condition is called a local Lipschitz condition, which is much stronger than continuity, and a bit weaker than differentiability. – Thomas Apr 27 '23 at 19:41

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As stated in a comment, the property that you define is called "local Lipschitz continuity" and is much stronger than continuity. Indeed, you are requiring some form of quantification of the rate at which $f(a+h)$ approaches $f(a)$ when $h \to 0$. More precisely you ask that the convergence be at most linear. Continuity just requires that $f(a+h) \to f(a)$, without prescribing the rate at which this must hold.

A typical counter-example would be the square-root function: $$ f: \begin{cases} \mathbb{R}_+ \to \mathbb{R}_+, \\ x \mapsto \sqrt{x}. \end{cases} $$

Take $a = 0$. Hence, for any $h > 0$, $f(a+h) - f(a) = \sqrt{h}$. You can check that for any $M, N \in \mathbb{N}^*$, there exists $0 < h < \frac 1 N$ such that $\sqrt{h} > M h$.

To be complete, the square-root function is called $\frac 12$-Hölder continuous, meaning exactly that the following quantification holds: $$ |f(a+h)-f(a)| \leq M |h|^{\frac 12}. $$

As you imagine, for any $\alpha \in (0,1)$, one can define a notion of $\alpha$-Hölder continuous, where the bound is $|h|^\alpha$. All these notions are strictly stronger than continuity, but weaker than Lipschitz continuity.

All these tools help to qualify the regularity of the functions.

cs89
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  • Thank you for the detailed explanation. It was most helpful. Do you know if it is true that every continuous function is $\alpha$-Holder continuous for some $\alpha$? I’m guessing no. – skh Apr 27 '23 at 20:19
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    You guessed right. Hölder continuity is also a form of quantification of continuity (which is an unquantified property). Continuous functions can be continuous at arbitrarily bad rates. For example, you can check that $f(x) := -1/\ln(x)$ for $x > 0$ extended by $f(0) = 0$ is continuous on $[0,+\infty)$, but not $\alpha$-Hölder continuous for any $\alpha > 0$. – cs89 Apr 27 '23 at 20:56